Define oxidation and reduction in terms of electron transfer and change in oxidation number.

Short Answer

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Oxidation is the loss of electrons (increase in oxidation number); reduction is the gain of electrons (decrease in oxidation number).

Step by step solution

01

Understanding Oxidation in Terms of Electron Transfer

Oxidation is the process where an atom or molecule loses electrons during a chemical reaction. For example, when magnesium reacts with oxygen, it loses two electrons and forms a cation: \[ \text{Mg} \rightarrow \text{Mg}^{2+} + 2e^- \] This loss of electrons signifies that magnesium has been oxidized.
02

Understanding Reduction in Terms of Electron Transfer

Reduction is the process where an atom or molecule gains electrons during a chemical reaction. Continuing with the earlier example, when oxygen gains the electrons lost by magnesium, it forms an anion: \[ \text{O}_2 + 4e^- \rightarrow 2\text{O}^{2-} \] This gain of electrons signifies that oxygen has been reduced.
03

Defining Oxidation in Terms of Oxidation Number

Oxidation can also be described in terms of an increase in oxidation number. For example, in the reaction of magnesium with oxygen, the oxidation number of magnesium increases from 0 to +2: \[ \text{Mg} (0) \rightarrow \text{Mg}^{2+} (+2) \] This increase in oxidation number indicates oxidation.
04

Defining Reduction in Terms of Oxidation Number

Reduction can be described in terms of a decrease in oxidation number. For example, in the same reaction, the oxidation number of oxygen decreases from 0 to -2: \[ \text{O}_2 (0) \rightarrow 2\text{O}^{2-} (-2) \] This decrease in oxidation number indicates reduction.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electron Transfer
Electron transfer is a fundamental concept in chemistry, especially when discussing redox reactions. To break it down, let's look at a simple reaction between magnesium and oxygen. This reaction involves the transfer of electrons from magnesium to oxygen.

In this example, magnesium starts as a neutral atom. During the oxidation process, it loses two electrons, which transforms it into a magnesium ion with a +2 charge \( \text{Mg} \rightarrow \text{Mg}^{2+} + 2e^- \).
On the other hand, oxygen gains those lost electrons during the reduction process. Oxygen molecules each gain two electrons to form an oxide ion with a -2 charge \( \text{O}_2 + 4e^- \rightarrow 2\text{O}^{2-} \), thus completing the electron transfer.

The takeaway here is that:*Oxidation* involves *losing electrons,* while
*Reduction* involves *gaining electrons.* Understanding this electron transfer is key to grasping how redox reactions work.
Oxidation Number
Oxidation numbers are used to keep track of electrons in atoms during a chemical reaction. They help us determine which atoms are oxidized and which are reduced.

Oxidation is defined as an increase in oxidation number. Let’s revisit the magnesium and oxygen reaction. Initially, the oxidation number of magnesium is 0 but after oxidation, it changes to +2 \( \text{Mg} (0) \rightarrow \text{Mg}^{2+} (+2) \).

Reduction, on the other hand, is defined by a decrease in oxidation number. In our example, oxygen starts with an oxidation number of 0 and it decreases to -2 upon gaining electrons \( \text{O}_2 (0) \rightarrow 2\text{O}^{2-} (-2) \).

In summary, recognizing changes in oxidation numbers helps us identify which elements are gaining or losing electrons in a reaction.
Redox Reactions
Redox reactions (short for reduction-oxidation reactions) are chemical processes that involve the transfer of electrons between two substances. Each reaction involves both a reduction process and an oxidation process, making them paired.

For instance, in the reaction between magnesium and oxygen we've been discussing, magnesium is oxidized \(\text{losing electrons}\), while oxygen is reduced \(\text{gaining electrons}\). These two processes are intertwined; one cannot occur without the other.

In a typical redox reaction:
  • The *oxidizing agent* is the substance that gets reduced (gains electrons).
  • The *reducing agent* is the substance that gets oxidized (loses electrons).
Understanding these roles can help clarify how redox reactions work and why they are essential in both natural processes and industrial applications.

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Most popular questions from this chapter

Both a D-sized and an AAA-sized alkaline battery have an output of \(1.5 \mathrm{~V}\). What property of the cell potential allows this to occur? What is different about these two batteries?

Use the following half-reactions to write three spontaneous reactions, calculate \(E_{\text {cell }}^{\circ}\) for each reaction, and rank the strengths of the oxidizing and reducing agents: (1) \(\mathrm{Al}^{3+}(a q)+3 \mathrm{e}^{-} \longrightarrow \mathrm{Al}(s) \quad E^{\circ}=-1.66 \mathrm{~V}\) (2) \(\mathrm{N}_{2} \mathrm{O}_{4}(g)+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{NO}_{2}^{-}(a q) \quad E^{\circ}=0.867 \mathrm{~V}\) (3) \(\mathrm{SO}_{4}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(I)+2 \mathrm{e}^{-} \longrightarrow \mathrm{SO}_{3}^{2-}(a q)+2 \mathrm{OH}^{-}(a q)\) \(E^{\circ}=0.93 \mathrm{~V}\)

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A chemist designs an ion-specific probe for measuring \(\left[\mathrm{Ag}^{+}\right]\) in an \(\mathrm{NaCl}\) solution saturated with \(\mathrm{AgCl}\). One half-cell has an Ag wire electrode immersed in the unknown AgCl-saturated \(\mathrm{NaCl}\) solution. It is connected through a salt bridge to the other half-cell, which has a calomel reference electrode [a platinum wire immersed in a paste of mercury and calomel (Hg \(_{2} \mathrm{Cl}_{2}\) ) ] in a saturated KCl solution. The measured \(E_{\text {cell }}\) is \(0.060 \mathrm{~V}\). (a) Given the following standard half-reactions, calculate \(\left[\mathrm{Ag}^{+}\right]\). Calomel: \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}(s)+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Hg}(I)+2 \mathrm{Cl}^{-}(a q) \quad E^{\circ}=0.24 \mathrm{~V}\) Silver: \(\mathrm{Ag}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \operatorname{Ag}(s)\) $$E^{\circ}=0.80 \mathrm{~V}$$ (Hint: Assume that [Cl \(^{-}\) ] is so high that it is essentially constant.) (b) A mining engineer wants an ore sample analyzed with the \(\mathrm{Ag}^{+}-\) selective probe. After pretreating the ore sample, the chemist measures the cell voltage as \(0.53 \mathrm{~V}\). What is \(\left[\mathrm{Ag}^{+}\right] ?\)

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